$\mathrm{K}_{\mathrm{c}}$ for the reaction, $\mathrm{A}_2(\mathrm{~g}) \stackrel{\mathrm{T}(\mathrm{K})}{\rightleftarrows} \mathrm{B}_2(\mathrm{~g})$ is 99.0 . In a 1 L closed flask two moles of $\mathrm{B}_2(\mathrm{~g})$ is heated to $\mathrm{T}(\mathrm{K})$. What is the concentration of…

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$\mathrm{K}_{\mathrm{c}}$ for the reaction, $\mathrm{A}_2(\mathrm{~g}) \stackrel{\mathrm{T}(\mathrm{K})}{\rightleftarrows} \mathrm{B}_2(\mathrm{~g})$ is 99.0 . In a 1 L closed flask two moles of $\mathrm{B}_2(\mathrm{~g})$ is heated to $\mathrm{T}(\mathrm{K})$. What is the concentration of $\mathrm{B}_2(\mathrm{~g})$ (in $\mathrm{mol} \mathrm{L}^{-1}$ ) at equilibrium?

Aicharya · Accepted Answer

Initial concentration of $\mathrm{A}_2=0$ Initial concentration of $\mathrm{B}_2=2$ Let x be the concentration of $\mathrm{A}_2$ so the change in $\mathrm{B}_2$ would be $-2 x$ Equilibrium concentration are $\left[\mathrm{A}_2\right]=\mathrm{x}, \quad\left[\mathrm{B}_2\right]=2-2 \mathrm{x}$ Putting the value in eq. (1) $99=\frac{(2-2 x)^2}{x} \Rightarrow 99=4 \frac{(1-x)^2}{x}$ Let's simplify the equation $-\left[(\mathrm{a}-\mathrm{b})^2=\mathrm{a}^2-2 \mathrm{ab}+\mathrm{b}^2\right]$ $99=\frac{4\left(1-2 x+x^2\right)}{x}$ $99 \mathrm{x}=4-8 \mathrm{x}+4 \mathrm{x}^2 \Rightarrow 4 \mathrm{x}^2-107 \mathrm{x}+4=0$ Solving the quadratic equation we use the Quadratic formula:- $\mathrm{x}=\frac{-\mathrm{b} \pm \sqrt{\mathrm{b}^2-4 \mathrm{ac}}}{2 \mathrm{a}} \quad \mathrm{a}=4, \mathrm{~b}=-107, \mathrm{c}=4$ $x=\frac{-(-107) \pm \sqrt{(-107)^2-4(4)(4)}}{2 \times 4}$ $x=\frac{107 \pm \sqrt{11449-64}}{8} \Rightarrow x=\frac{107 \pm \sqrt{11385}}{8}$ $x=\frac{107 \pm 106.70}{8} \Rightarrow x=\frac{213.65}{8} \approx 26.71$ $x=\frac{0.30}{x} \approx 0.030$ Putting the value of $x$ to finding the concentration of $\mathrm{B}_2$ $B_2=2-2 x \Rightarrow B_2=2-2 \times 0.03$ $B_2=2=0.06=1.9$

$\mathrm{K}_{\mathrm{c}}$ for the reaction, $\mathrm{A}_2(\mathrm{~g})…

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